| ⇦ | 
| ⇨ |
A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 sec in earth’s horizontal magnetic field of 24 microtesla. When a horizontal field of 18 microtesla is produced opposite to the earth’s field by placing a current carrying wire, the new time period of magnet will be
Options
(a) 1 s
(b) 2 s
(c) 3 s
(d) 4 s
Correct Answer:
4 s
Explanation:
Time period of a vibration magnetometer,
T 1 / √B
T₁ / T₂ = √(B₂ / B₁)
T₂ = T₁ √(B₁ / B₂)
2 √ (24 x 10⁻⁶ / 6 x 10⁻⁶) = 4 sec.
Related Questions: - An alternating emf given by equation
e=300 sin[(100 π)t] V
- A Gaussian surface in the cylinder of cross-section πa² and length L is immersed
- When a mass M is attached to a spring of force constant K,
- An A.C source is rated at 220 V, 50 Hz. The time taken for voltage to change
- Bragg’s law for X-rays is
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An alternating emf given by equation e=300 sin[(100 π)t] V
- A Gaussian surface in the cylinder of cross-section πa² and length L is immersed
- When a mass M is attached to a spring of force constant K,
- An A.C source is rated at 220 V, 50 Hz. The time taken for voltage to change
- Bragg’s law for X-rays is
Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply